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Bertrand's Theorem says: the only forces whose bounded orbits imply closed orbits are the Hooke's law and the attractive inverse square force.

I'm looking at the Hooke's law $f=-k r$ and try to see explicitly that the orbit is indeed closed.

I use the orbit equation $$\frac{d^{2} u}{d \theta^{2}}+u=\frac{-m}{l^{2} u^{2}} f\left(\frac{1}{u}\right)$$ with the force given as $f=-k r$, therefore I get $$\frac{d^{2} u}{d \theta^{2}}+u=+\frac{mk}{l^{2} u^3}$$ as the equation defining the trajectory.

However neither can I solve this nor can I see that the equation implies a closed orbit.

Can you please help me.

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    $\begingroup$ The Wikipedia article on Bertrand's theorem includes a proof. $\endgroup$
    – John Rennie
    Commented Nov 2, 2021 at 7:26
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    $\begingroup$ Terminology: Hooke's law refers to the deformation of an elastic body such as a spring or wire. (a) You can indeed represent Hooke's law by $\vec F =-k \vec r$, but note that $\vec r$ would be the extension of the body, not the distance of the free end of the spring from the tethered end. (b) In your context, $\vec F =-k \vec r$ is an abstract relationship: it need not arise from deformation of an elastic body. I'd advise against calling it 'Hooke's law'. $\endgroup$
    – Philip Wood
    Commented Nov 2, 2021 at 10:36
  • $\begingroup$ @Philip wood, I'm just copying Goldsteins terminology but your point is valid. $\endgroup$
    – Kashmiri
    Commented Nov 2, 2021 at 12:25
  • $\begingroup$ @John rennie. Thank you. Got it. $\endgroup$
    – Kashmiri
    Commented Nov 2, 2021 at 12:27
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    $\begingroup$ @Kasmiri Interesting. Thank you. But I acknowledge that Goldstein does have redeeming features! $\endgroup$
    – Philip Wood
    Commented Nov 2, 2021 at 13:43

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As @JohnRennie notes, it's best not to solve this with the Binet equation. Since $\vec{f}=-k\vec{r}$,$$m\ddot{x}_i=-kx_i,$$which is just simple harmonic motion. In particular, each Cartesian coordinate $x_i$ oscillates with period $\frac{2\pi}{\omega}$ where $\omega:=\sqrt{\frac{k}{m}}$.

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  • $\begingroup$ And heck, if I was looking to solve the radial EOM, I would start by solving the SHM Cartesian components, transforming to polar coordinates, and checking that the resulting function satisfies the desired ODE. It's not automatic - it's not obvious that the re-expression as $r=r(\theta)$ will work cleanly - but it's well worth a try at the start. $\endgroup$
    – Emilio Pisanty
    Commented Nov 2, 2021 at 10:28
  • $\begingroup$ @EmilioPisanty Good point. That amounts to checking the SHM solutions conserve angular momentum. (Of course they all do, because the force is radial.) $\endgroup$
    – J.G.
    Commented Nov 2, 2021 at 10:31
  • $\begingroup$ Yeah, I guess it can be seen that way. It's not obvious to me that it will work (because getting $r=r(\theta)$ from $r=r(t)$ involves obtaining and inverting $\theta(t)$, which is not guaranteed to work) so there's probably nontrivial work involved, but if a core solution to the underlying ODE problem already exists then it's silly to ignore it. $\endgroup$
    – Emilio Pisanty
    Commented Nov 2, 2021 at 11:42

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